This is the first problem from a 2005 International Math Olympiad page. The description here only partially matches the picture, but I'll explain.
I encourage you to draw a picture of the hexagon described in that problem statement, because I am starting here by drawing what you see above instead. I set point \(\small\mathtt{B}\) of \(\small\Delta\mathtt{ABC}\) at the origin on a coordinate plane and draw ray \(\small\mathtt{BA}\) (\(\small\mathtt{A}\) is not pictured yet) at a 60\(^\circ\)angle to the \(\small\mathtt{x}\)-axis. Side \(\small\mathtt{BC}\) will lie on the \(\small\mathtt{x}\)-axis, and \(\small\measuredangle\mathtt{B}\) must be 60\(^\circ\)(same with \(\small\measuredangle\mathtt{A}\) and \(\small \measuredangle\mathtt{C}\)) because an equilateral triangle has 3 congruent angles. I've also marked a position for \(\small\mathtt{A_1}\), which I've assigned the coordinates (\(\small\mathtt{x_1 , 0}\)).
Let's make the first side of the regular hexagon, which has equal side lengths and equal angle measures. Each angle measure is \(\small\mathtt {\frac{6 - 2 \cdot 180}{6}}\), or 120\(^\circ\). I can shoot a line at a 120\(^\circ\)angle toward the \(\small\mathtt{y}\)-axis to form our first side, which I'll call \(\small\mathtt{s}\) (it's \(\small\overline{\mathtt{A_{1}C_2}}\) in the picture).
\(\small\Delta\mathtt{PBA_1}\) is a 30-60-90 triangle, which tells us that the coordinates of point \(\small\mathtt{P}\) are (\(\small\mathtt{0, \sqrt{3}x_1}\)). Line \(\small\mathtt{PA_1}\), then, has the equation \(\small\mathtt{y = -\sqrt{3}x + \sqrt{3}x_1}\), and the line that will form side \(\small\mathtt{AB}\) of the triangle has the equation \(\small\mathtt{\sqrt{3}x}\). The intersection of these lines, point \(\small\mathtt{C_2}\), is then at (\(\small\mathtt{\frac{x_1}{2}, \frac{\sqrt{3}x_1}{2}}\)).
Finally, we can run a vertical line up from \(\small\mathtt{A_1}\) to find \(\small\mathtt{C_1}\). From \(\small\mathtt{C_2}\), you can walk \(\small\mathtt{\frac{x_1}{2}}\) to the right to get to the same horizontal position as \(\small\mathtt{A_1}\) and up a distance of \(\small\mathtt {\frac{\sqrt{3}x_1}{2}}\) to get to the same vertical position as \(\small\mathtt{P}\). This is the exact same walk we would make from point \(\small \mathtt{B}\) to get to \(\small\mathtt{C_2}\), so we know that we're still on the side of the triangle. And it's the same walk in terms of distances that one would make from point \(\small\mathtt{A_1}\) to \(\small\mathtt{C_2}\), so we know that \(\small\mathtt{C_{1}C{2}}\) has length \(\small \mathtt{s}\). This means we also know the coordinates of \(\small\mathtt{C_1}\): (\(\small\mathtt{x_1, \sqrt{3}x_1}\)). It's probably also worth pointing out that \(\small\mathtt{x_1 = s}\). The equilateral triangle \(\small\Delta\mathtt{A_{1}BC_2}\) shows us this.
Bridge with the Givens
Now we can use the givens to fill in the picture. And symmetry. That concept alone will save me paragraphs of writing. Since we can construct the right-hand side of this entire image using symmetry, I am able to easily give you the handy table listing the coordinates of all the points of the hexagon shown below.
| Point | Coordinates |
|---|---|
| \(\small\mathtt{A_1}\) | (\(\small\mathtt{x_1, 0}\)) |
| \(\small\mathtt{A_2}\) | (\(\small\mathtt{x_1 + s, 0}\)) |
| \(\small\mathtt{B_1}\) | (\(\small\mathtt{\frac{3x_1}{2} + s, \frac{\sqrt{3}x_1}{2}}\)) |
| \(\small\mathtt{B_2}\) | (\(\small\mathtt{x_1 + s, \sqrt{3}x_1}\)) |
| \(\small\mathtt{C_1}\) | (\(\small\mathtt{x_1, \sqrt{3}x_1}\)) |
| \(\small\mathtt{C_2}\) | (\(\small\mathtt{\frac{x_1}{2}, \frac{\sqrt{3}x_1}{2}}\)) |
What remains is for us to show that the lines \(\small\mathtt{A_1 B_2}\), \(\small\mathtt{B_1 C_2}\), and \(\small\mathtt{C_1 A_2}\) are concurrent, or that they all meet at a point. And now that we have the coordinates of the vertices, we are left with the task of solving a system of 3 equations—or at least the task of showing that a system of 3 equations has a solution. I think we actually have too much information now, but that gives a lot of options.
Symmetry and Now Parallelisms FTW!
Since \(\small\mathtt{x_1 = s}\), a walk from point \(\small\mathtt{B}\) to \(\small\mathtt{A_1}\) and then up to \(\small\mathtt{C_1}\) is the same as a walk from \(\small\mathtt{A_1}\) to \(\small\mathtt{A_2}\) and then up to \(\small\mathtt{B_2}\). So, lines \(\small\mathtt{AB}\) and \(\small \mathtt{A_{1}B_2}\) are parallel. Thus, the equation for line \(\small\mathtt{A_{1}B_2}\) is the same as the one for line \(\small\mathtt{AB}\) (\(\small\mathtt{y = \sqrt{3}x}\)) but just shifted to the right \(\small\mathtt{x_1}\), or \(\small\mathtt{s}\), units: \(\small\mathtt{y = \sqrt{3}(x - x_1)}\).
Similarly, the equation for line \(\small\mathtt{C_{1}A_2}\) is the same as the one for line \(\small\mathtt{C_{2}A_1}\) (\(\small\mathtt{y = - \sqrt{3}x + \sqrt{3}x_1}\)) but shifted to the right \(\small\mathtt{x_1}\), or \(\small\mathtt{s}\), units: \(\small\mathtt{y = -\sqrt{3}(x - x_1) + \sqrt{3}x_1}\).
Finally, the horizontal line \(\small\mathtt{B_{1}C_2}\) has the equation \(\small\mathtt{y = \frac{\sqrt{3}x_1}{2}}\).
If you can slog through that algebra, the point of intersection of those three lines is at (\(\small\mathtt{\frac{3s}{2}, \frac{\sqrt{3}s}{2}}\)). Whew!